Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 7.6.2

Interactive: Dielectric Slab and Optical Fibre

Problem 7.6.2

Surface waves on a coated conductor

A perfectly conducting plane is coated with a lossless dielectric layer, $ε_{r} = 5$ , $μ = μ_{0}$ , of thickness $t = 1 cm$ . Above it is air. Surface waves of $f = 10 GHz$ travel along $z$ .

Which surface-wave modes propagate, and what are their cutoff frequencies?
For the TM $_{0}$ mode find $τ$ , $n_{eff}$ and $α$ , and the height above the coating at which the field has dropped by $20 dB$ .
Find the surface impedance $Z_{s} = - E_{z} / H_{x}$ seen from the air. Is it inductive or capacitive?
Up to which thickness does the coating guide TM $_{0}$ alone at $10 GHz$ ?

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.10-36.

Try it yourself first.

Strategy

Image theory (Section 7.6.9): the metal plane can replace the midplane $y = 0$ of a slab of thickness $d = 2 t$ for every slab mode whose tangential electric field already vanishes there. These are the odd TE $_{m}$ ( $E_{x} \propto sin h y$ ) and the even TM $_{m}$ ( $E_{z} \propto d H_{x} / d y \propto sin h y$ ). The $2 cm$ slab is that of Problem 7.6.1, so its roots can be reused.

1. Bookkeeping

$n_{1}^{2} = 5$ , $n_{2} = 1$ , $t = 0.01 m$ , $k_{0} = 209.4 rad/m$ and $ω ε_{0} = 2 π \times 1 0^{10} \times 1 0^{- 9} / (36 π) = 0.5556 S/m$ .

$V = k_{0} t NA = 209.4 \times 0.01 \times 2 = 4.189$ , as in Problem 7.6.1. The cutoffs of the slab modes are $f_{c, m} = m c / (4 t NA) = 3.75 m GHz$ , and the coated plane keeps only

TM_{0} (f_{c} = 0), TE_{1} (3.75 GHz), TM_{2} (7.5 GHz), TE_{3} (11.25 GHz), \dots

At $10 GHz$ TM $_{0}$ , TE $_{1}$ and TM $_{2}$ propagate. TE $_{0}$ , TM $_{1}$ and TE $_{2}$ of the full slab are absent: they would need $E_{tan} \neq = 0$ on the metal.

TM $_{0}$ : $w = (n_{2}^{2} / n_{1}^{2}) u tan u = 0.2 u tan u$ with $u^{2} + w^{2} = 4.18 9^{2}$ gives $u = 1.495$ , $w = 3.913$ . With $t$ in place of $d /2$ ,

h α τ n_{eff} = \frac{u}{t} = 149.5 rad/m, = \frac{w}{t} = 391.3 Np/m, = 5 \times 209. 4^{2} - 149. 5^{2} = 443.8 rad/m, = \frac{443.8}{209.4} = 2.119.

Above the coating $H_{x} \propto e^{- α s}$ , $s$ the height above the surface. $20 dB$ means $e^{- α s} = 0.1$ : $s = ln 10/ α = 5.9 mm$ .

In the air $E_{z} = \frac{j}{ω ε _{0}} \frac{d H _{x}}{d y} = \frac{j}{ω ε _{0}} (- α) H_{x}$ , so

Z_{s} = - \frac{E _{z}}{H _{x}} = j \frac{α}{ω ε _{0}} = j \frac{391.3}{0.5556} = j 704 Ω.

A positive reactance: the surface is inductive.

TE $_{1}$ must be below cutoff: $k_{0} t NA < π /2$ , so

t < \frac{λ _{0}}{4 NA} = \frac{3 cm}{8} = 3.75 mm .

Answer

TM_{0}, TE_{1}, TM_{2}; t < 3.75 mm Z_{s} = j 704 Ω TM_{0} : τ = 443.8 rad/m, n_{eff} = 2.119, α = 391.3 Np/m, s_{20 dB} = 5.9 mm

Only the odd TE and even TM branches survive the metal below. Read $Z_{s} = j 704 Ω$ of TM $_{0}$ .

InteractiveLoading simulation…

Check

$0.2 \times 1.495 \times tan 1.495 = 0.2 \times 1.495 \times 13.09 = 3.913 = w$ . $n_{eff} = 2.119$ is the TM $_{0}$ value of the $2 cm$ slab in the remark of Problem 7.6.1, as image theory says it must be.

(a) TM_0 profiles at 10\,\mathrm{GHz}. — (a) TM $_{0}$ profiles at $10 GHz$ .

(b) n_{\rm eff} of the three modes. — (a) TM $_{0}$ profiles at $10 GHz$ .

Remark

A TE surface wave would see a capacitive $Z_{s}$ , Eq. (7.6.33). A metal surface made inductive by a coating (or by corrugations) is the classic way to bind a TM surface wave, for surface-wave antennas and launchers. Exercise 7.6.3 runs this problem backwards: it prescribes the reactance and asks for the coating.

Problem 7.6.2: Surface waves on a coated conductor